Yes, that's a nice description. On the other hand P(S) can have a nest of length |P(S)|.

> The most familiar special case is that P(omega), although it contains > uncountable chains ordered like the reals, contains no chain of type > omega_1 or omega_1^*.

> > Can the push down lemma be extended to show f is eventually constant?

> Yes, easily, if kappa is regular; no, if kappa is singular. Suppose,> e.g., that eta = omega and S = omega. Your descending function> f:omega_{omega} -> P(omega) cannot be injective, but neither does it> have to be eventually constant. For examply, you could have f(mu) = S> for the first aleph_0 values of mu, f(mu) = S\{0| for the next aleph_1> values, f(mu) = S\{0,1} for the next aleph_2 values, and so on.